Exponents and Scientific Notation
Learning Outcomes
- Use the rules of exponents to simplify exponential expressions.
- Use scientific notation.
Rules for Exponents
Consider the product [latex]{x}^{3}\cdot {x}^{4}[/latex]. Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression.A General Note: The Product Rule of Exponents
For any real number [latex]a[/latex] and natural numbers [latex]m[/latex] and [latex]n[/latex], the product rule of exponents states thatExample: Using the Product Rule
Write each of the following products with a single base. Do not simplify further.- [latex]{t}^{5}\cdot {t}^{3}[/latex]
- [latex]\left(-3\right)^{5}\cdot \left(-3\right)[/latex]
- [latex]{x}^{2}\cdot {x}^{5}\cdot {x}^{3}[/latex]
Answer: Use the product rule to simplify each expression.
- [latex]{t}^{5}\cdot {t}^{3}={t}^{5+3}={t}^{8}[/latex]
- [latex]{\left(-3\right)}^{5}\cdot \left(-3\right)={\left(-3\right)}^{5}\cdot {\left(-3\right)}^{1}={\left(-3\right)}^{5+1}={\left(-3\right)}^{6}[/latex]
- [latex]{x}^{2}\cdot {x}^{5}\cdot {x}^{3}[/latex]
Try It
Write each of the following products with a single base. Do not simplify further.- [latex]{k}^{6}\cdot {k}^{9}[/latex]
- [latex]{\left(\dfrac{2}{y}\right)}^{4}\cdot \left(\dfrac{2}{y}\right)[/latex]
- [latex]{t}^{3}\cdot {t}^{6}\cdot {t}^{5}[/latex]
Answer:
- [latex]{k}^{15}[/latex]
- [latex]{\left(\dfrac{2}{y}\right)}^{5}[/latex]
- [latex]{t}^{14}[/latex]
Using the Quotient Rule of Exponents
The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but different exponents. In a similar way to the product rule, we can simplify an expression such as [latex]\dfrac{{y}^{m}}{{y}^{n}}[/latex], where [latex]m>n[/latex]. Consider the example [latex]\dfrac{{y}^{9}}{{y}^{5}}[/latex]. Perform the division by canceling common factors.A General Note: The Quotient Rule of Exponents
For any real number [latex]a[/latex] and natural numbers [latex]m[/latex] and [latex]n[/latex], such that [latex]m>n[/latex], the quotient rule of exponents states thatExample: Using the Quotient Rule
Write each of the following products with a single base. Do not simplify further.- [latex]\dfrac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}[/latex]
- [latex]\dfrac{{t}^{23}}{{t}^{15}}[/latex]
- [latex]\dfrac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}[/latex]
Answer: Use the quotient rule to simplify each expression.
- [latex]\dfrac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}={\left(-2\right)}^{14 - 9}={\left(-2\right)}^{5}[/latex]
- [latex]\dfrac{{t}^{23}}{{t}^{15}}={t}^{23 - 15}={t}^{8}[/latex]
- [latex]\dfrac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}={\left(z\sqrt{2}\right)}^{5 - 1}={\left(z\sqrt{2}\right)}^{4}[/latex]
Try It
Write each of the following products with a single base. Do not simplify further.- [latex]\dfrac{{s}^{75}}{{s}^{68}}[/latex]
- [latex]\dfrac{{\left(-3\right)}^{6}}{-3}[/latex]
- [latex]\dfrac{{\left(e{f}^{2}\right)}^{5}}{{\left(e{f}^{2}\right)}^{3}}[/latex]
Answer:
- [latex]{s}^{7}[/latex]
- [latex]{\left(-3\right)}^{5}[/latex]
- [latex]{\left(e{f}^{2}\right)}^{2}[/latex]
Using the Power Rule of Exponents
Suppose an exponential expression is raised to some power. Can we simplify the result? Yes. To do this, we use the power rule of exponents. Consider the expression [latex]{\left({x}^{2}\right)}^{3}[/latex]. The expression inside the parentheses is multiplied twice because it has an exponent of 2. Then the result is multiplied three times because the entire expression has an exponent of 3.Product Rule | Power Rule | |||||||||
---|---|---|---|---|---|---|---|---|---|---|
[latex]5^{3}\cdot5^{4}[/latex] | = | [latex]5^{3+4}[/latex] | = | [latex]5^{7}[/latex] | but | [latex]\left(5^{3}\right)^{4}[/latex] | = | [latex]5^{3\cdot4}[/latex] | = | [latex]5^{12}[/latex] |
[latex]x^{5}\cdot x^{2}[/latex] | = | [latex]x^{5+2}[/latex] | = | [latex]x^{7}[/latex] | but | [latex]\left(x^{5}\right)^{2}[/latex] | = | [latex]x^{5\cdot2}[/latex] | = | [latex]x^{10}[/latex] |
[latex]\left(3a\right)^{7}\cdot\left(3a\right)^{10} [/latex] | = | [latex]\left(3a\right)^{7+1-} [/latex] | = | [latex]\left(3a\right)^{17}[/latex] | but | [latex]\left(\left(3a\right)^{7}\right)^{10} [/latex] | = | [latex]\left(3a\right)^{7\cdot10} [/latex] | = | [latex]\left(3a\right)^{70}[/latex] |
A General Note: The Power Rule of Exponents
For any real number [latex]a[/latex] and positive integers [latex]m[/latex] and [latex]n[/latex], the power rule of exponents states thatExample: Using the Power Rule
Write each of the following products with a single base. Do not simplify further.- [latex]{\left({x}^{2}\right)}^{7}[/latex]
- [latex]{\left({\left(2t\right)}^{5}\right)}^{3}[/latex]
- [latex]{\left({\left(-3\right)}^{5}\right)}^{11}[/latex]
Answer: Use the power rule to simplify each expression.
- [latex]{\left({x}^{2}\right)}^{7}={x}^{2\cdot 7}={x}^{14}[/latex]
- [latex]{\left({\left(2t\right)}^{5}\right)}^{3}={\left(2t\right)}^{5\cdot 3}={\left(2t\right)}^{15}[/latex]
- [latex]{\left({\left(-3\right)}^{5}\right)}^{11}={\left(-3\right)}^{5\cdot 11}={\left(-3\right)}^{55}[/latex]
Try It
Write each of the following products with a single base. Do not simplify further.- [latex]{\left({\left(3y\right)}^{8}\right)}^{3}[/latex]
- [latex]{\left({t}^{5}\right)}^{7}[/latex]
- [latex]{\left({\left(-g\right)}^{4}\right)}^{4}[/latex]
Answer:
- [latex]{\left(3y\right)}^{24}[/latex]
- [latex]{t}^{35}[/latex]
- [latex]{\left(-g\right)}^{16}[/latex]
Zero and Negative Exponents
Return to the quotient rule. We made the condition that [latex]m>n[/latex] so that the difference [latex]m-n[/latex] would never be zero or negative. What would happen if [latex]m=n[/latex]? In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example.[latex]\dfrac{t^{8}}{t^{8}}=\dfrac{\cancel{t^{8}}}{\cancel{t^{8}}}=1[/latex]
If we were to simplify the original expression using the quotient rule, we would haveA General Note: The Zero Exponent Rule of Exponents
For any nonzero real number [latex]a[/latex], the zero exponent rule of exponents states thatExample: Using the Zero Exponent Rule
Simplify each expression using the zero exponent rule of exponents.- [latex]\dfrac{{c}^{3}}{{c}^{3}}[/latex]
- [latex]\dfrac{-3{x}^{5}}{{x}^{5}}[/latex]
- [latex]\dfrac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}[/latex]
- [latex]\dfrac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}[/latex]
Answer: Use the zero exponent and other rules to simplify each expression.
- [latex]\begin{align}\frac{c^{3}}{c^{3}} & =c^{3-3} \\ & =c^{0} \\ & =1\end{align}[/latex]
- [latex]\begin{align} \frac{-3{x}^{5}}{{x}^{5}}& = -3\cdot \frac{{x}^{5}}{{x}^{5}} \\ & = -3\cdot {x}^{5 - 5} \\ & = -3\cdot {x}^{0} \\ & = -3\cdot 1 \\ & = -3 \end{align}[/latex]
- [latex]\begin{align} \frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}& = \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{1+3}} && \text{Use the product rule in the denominator}. \\ & = \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{4}} && \text{Simplify}. \\ & = {\left({j}^{2}k\right)}^{4 - 4} && \text{Use the quotient rule}. \\ & = {\left({j}^{2}k\right)}^{0} && \text{Simplify}. \\ & = 1 \end{align}[/latex]
- [latex]\begin{align} \frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}& = 5{\left(r{s}^{2}\right)}^{2 - 2} && \text{Use the quotient rule}. \\ & = 5{\left(r{s}^{2}\right)}^{0} && \text{Simplify}. \\ & = 5\cdot 1 && \text{Use the zero exponent rule}. \\ & = 5 && \text{Simplify}. \end{align}[/latex]
Try It
Simplify each expression using the zero exponent rule of exponents.- [latex]\dfrac{{t}^{7}}{{t}^{7}}[/latex]
- [latex]\dfrac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}}[/latex]
- [latex]\dfrac{{w}^{4}\cdot {w}^{2}}{{w}^{6}}[/latex]
- [latex]\dfrac{{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}[/latex]
Answer:
- [latex]1[/latex]
- [latex]\dfrac{1}{2}[/latex]
- [latex]1[/latex]
- [latex]1[/latex]
Using the Negative Rule of Exponents
Another useful result occurs if we relax the condition that [latex]m>n[/latex] in the quotient rule even further. For example, can we simplify [latex]\dfrac{{h}^{3}}{{h}^{5}}[/latex]? When [latex]m<n[/latex]—that is, where the difference [latex]m-n[/latex] is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal. Divide one exponential expression by another with a larger exponent. Use our example, [latex]\dfrac{{h}^{3}}{{h}^{5}}[/latex].A General Note: The Negative Rule of Exponents
For any nonzero real number [latex]a[/latex] and natural number [latex]n[/latex], the negative rule of exponents states thatExample: Using the Negative Exponent Rule
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.- [latex]\dfrac{{\theta }^{3}}{{\theta }^{10}}[/latex]
- [latex]\dfrac{{z}^{2}\cdot z}{{z}^{4}}[/latex]
- [latex]\dfrac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}[/latex]
Answer:
- [latex]\dfrac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3 - 10}={\theta }^{-7}=\dfrac{1}{{\theta }^{7}}[/latex]
- [latex]\dfrac{{z}^{2}\cdot z}{{z}^{4}}=\dfrac{{z}^{2+1}}{{z}^{4}}=\dfrac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\dfrac{1}{z}[/latex]
- [latex]\dfrac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}={\left(-5{t}^{3}\right)}^{4 - 8}={\left(-5{t}^{3}\right)}^{-4}=\dfrac{1}{{\left(-5{t}^{3}\right)}^{4}}[/latex]
Try It
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.- [latex]\frac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}[/latex]
- [latex]\frac{{f}^{47}}{{f}^{49}\cdot f}[/latex]
- [latex]\frac{2{k}^{4}}{5{k}^{7}}[/latex]
Answer:
- [latex]\frac{1}{{\left(-3t\right)}^{6}}[/latex]
- [latex]\frac{1}{{f}^{3}}[/latex]
- [latex]\frac{2}{5{k}^{3}}[/latex]
Example: Using the Product and Quotient Rules
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.- [latex]{b}^{2}\cdot {b}^{-8}[/latex]
- [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}[/latex]
- [latex]\dfrac{-7z}{{\left(-7z\right)}^{5}}[/latex]
Answer:
- [latex]{b}^{2}\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\frac{1}{{b}^{6}}[/latex]
- [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}={\left(-x\right)}^{5 - 5}={\left(-x\right)}^{0}=1[/latex]
- [latex]\dfrac{-7z}{{\left(-7z\right)}^{5}}=\dfrac{{\left(-7z\right)}^{1}}{{\left(-7z\right)}^{5}}={\left(-7z\right)}^{1 - 5}={\left(-7z\right)}^{-4}=\dfrac{1}{{\left(-7z\right)}^{4}}[/latex]
Try It
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.- [latex]{t}^{-11}\cdot {t}^{6}[/latex]
- [latex]\dfrac{{25}^{12}}{{25}^{13}}[/latex]
Answer:
- [latex]{t}^{-5}=\dfrac{1}{{t}^{5}}[/latex]
- [latex]\dfrac{1}{25}[/latex]
Finding the Power of a Product
To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider [latex]{\left(pq\right)}^{3}[/latex]. We begin by using the associative and commutative properties of multiplication to regroup the factors.A General Note: The Power of a Product Rule of Exponents
For any real numbers [latex]a[/latex] and [latex]b[/latex] and any integer [latex]n[/latex], the power of a product rule of exponents states thatExample: Using the Power of a Product Rule
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.- [latex]{\left(a{b}^{2}\right)}^{3}[/latex]
- [latex]{\left(2t\right)}^{15}[/latex]
- [latex]{\left(-2{w}^{3}\right)}^{3}[/latex]
- [latex]\dfrac{1}{{\left(-7z\right)}^{4}}[/latex]
- [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}[/latex]
Answer: Use the product and quotient rules and the new definitions to simplify each expression.
- [latex]{\left(a{b}^{2}\right)}^{3}={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}={a}^{3}{b}^{6}[/latex]
- [latex]2{t}^{15}={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}[/latex]
- [latex]{\left(-2{w}^{3}\right)}^{3}={\left(-2\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}=-8\cdot {w}^{3\cdot 3}=-8{w}^{9}[/latex]
- [latex]\dfrac{1}{{\left(-7z\right)}^{4}}=\dfrac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\dfrac{1}{2,401{z}^{4}}[/latex]
- [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}={\left({e}^{-2}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}={e}^{-2\cdot 7}\cdot {f}^{2\cdot 7}={e}^{-14}{f}^{14}=\dfrac{{f}^{14}}{{e}^{14}}[/latex]
Try It
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.- [latex]{\left({g}^{2}{h}^{3}\right)}^{5}[/latex]
- [latex]{\left(5t\right)}^{3}[/latex]
- [latex]{\left(-3{y}^{5}\right)}^{3}[/latex]
- [latex]\dfrac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}[/latex]
- [latex]{\left({r}^{3}{s}^{-2}\right)}^{4}[/latex]
Answer:
- [latex]{g}^{10}{h}^{15}[/latex]
- [latex]125{t}^{3}[/latex]
- [latex]-27{y}^{15}[/latex]
- [latex]\dfrac{1}{{a}^{18}{b}^{21}}[/latex]
- [latex]\dfrac{{r}^{12}}{{s}^{8}}[/latex]
Finding the Power of a Quotient
To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following example.A General Note: The Power of a Quotient Rule of Exponents
For any real numbers [latex]a[/latex] and [latex]b[/latex] and any integer [latex]n[/latex], the power of a quotient rule of exponents states thatExample: Using the Power of a Quotient Rule
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.- [latex]{\left(\dfrac{4}{{z}^{11}}\right)}^{3}[/latex]
- [latex]{\left(\dfrac{p}{{q}^{3}}\right)}^{6}[/latex]
- [latex]{\left(\dfrac{-1}{{t}^{2}}\right)}^{27}[/latex]
- [latex]{\left({j}^{3}{k}^{-2}\right)}^{4}[/latex]
- [latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}[/latex]
Answer:
- [latex]{\left(\dfrac{4}{{z}^{11}}\right)}^{3}=\dfrac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}=\dfrac{64}{{z}^{11\cdot 3}}=\dfrac{64}{{z}^{33}}[/latex]
- [latex]{\left(\dfrac{p}{{q}^{3}}\right)}^{6}=\dfrac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}=\dfrac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}=\dfrac{{p}^{6}}{{q}^{18}}[/latex]
- [latex]{\left(\dfrac{-1}{{t}^{2}}\right)}^{27}=\dfrac{{\left(-1\right)}^{27}}{{\left({t}^{2}\right)}^{27}}=\dfrac{-1}{{t}^{2\cdot 27}}=\dfrac{-1}{{t}^{54}}=-\dfrac{1}{{t}^{54}}[/latex]
- [latex]{\left({j}^{3}{k}^{-2}\right)}^{4}={\left(\dfrac{{j}^{3}}{{k}^{2}}\right)}^{4}=\dfrac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}=\dfrac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}=\dfrac{{j}^{12}}{{k}^{8}}[/latex]
- [latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}={\left(\dfrac{1}{{m}^{2}{n}^{2}}\right)}^{3}=\dfrac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}=\dfrac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}=\dfrac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}=\dfrac{1}{{m}^{6}{n}^{6}}[/latex]
Try It
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.- [latex]{\left(\dfrac{{b}^{5}}{c}\right)}^{3}[/latex]
- [latex]{\left(\dfrac{5}{{u}^{8}}\right)}^{4}[/latex]
- [latex]{\left(\dfrac{-1}{{w}^{3}}\right)}^{35}[/latex]
- [latex]{\left({p}^{-4}{q}^{3}\right)}^{8}[/latex]
- [latex]{\left({c}^{-5}{d}^{-3}\right)}^{4}[/latex]
Answer:
- [latex]\dfrac{{b}^{15}}{{c}^{3}}[/latex]
- [latex]\dfrac{625}{{u}^{32}}[/latex]
- [latex]\dfrac{-1}{{w}^{105}}[/latex]
- [latex]\dfrac{{q}^{24}}{{p}^{32}}[/latex]
- [latex]\dfrac{1}{{c}^{20}{d}^{12}}[/latex]
Simplifying Exponential Expressions
Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.Example: Simplifying Exponential Expressions
Simplify each expression and write the answer with positive exponents only.- [latex]{\left(6{m}^{2}{n}^{-1}\right)}^{3}[/latex]
- [latex]{17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}[/latex]
- [latex]{\left(\dfrac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}[/latex]
- [latex]\left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)[/latex]
- [latex]{\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}[/latex]
- [latex]\dfrac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}[/latex]
Answer:
- [latex]\begin{align} {\left(6{m}^{2}{n}^{-1}\right)}^{3}& = {\left(6\right)}^{3}{\left({m}^{2}\right)}^{3}{\left({n}^{-1}\right)}^{3}&& \text{The power of a product rule} \\ & = {6}^{3}{m}^{2\cdot 3}{n}^{-1\cdot 3}&& \text{The power rule}\hfill \\ & = 216{m}^{6}{n}^{-3}&& \text{Simplify}. \\ & = \frac{216{m}^{6}}{{n}^{3}}&& \text{The negative exponent rule} \end{align}[/latex]
- [latex]\begin{align} {17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}& =& {17}^{5 - 4-3}&& \text{The product rule}\hfill \\ & = {17}^{-2}&& \text{Simplify}. \\ & = \frac{1}{{17}^{2}}\text{ or }\frac{1}{289}&& \text{The negative exponent rule} \end{align}[/latex]
- [latex]\begin{align} {\left(\frac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}& = \frac{{\left({u}^{-1}v\right)}^{2}}{{\left({v}^{-1}\right)}^{2}}&& \text{The power of a quotient rule} \\ & = \frac{{u}^{-2}{v}^{2}}{{v}^{-2}}&& \text{The power of a product rule} \\ & = {u}^{-2}{v}^{2-\left(-2\right)}&& \text{The quotient rule} \\ & = {u}^{-2}{v}^{4}&& \text{Simplify}. \\ & = \frac{{v}^{4}}{{u}^{2}}&& \text{The negative exponent rule} \end{align}[/latex]
- [latex]\begin{align} \left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)& =& -2\cdot 5\cdot {a}^{3}\cdot {a}^{-2}\cdot {b}^{-1}\cdot {b}^{2}&& \text{Commutative and associative laws of multiplication} \\ & = -10\cdot {a}^{3 - 2}\cdot {b}^{-1+2}&& \text{The product rule} \\ & = -10ab&& \text{Simplify}. \end{align}[/latex]
- [latex]\begin{align} {\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}& = {\left({x}^{2}\sqrt{2}\right)}^{4 - 4} && \text{The product rule} \\ & = {\left({x}^{2}\sqrt{2}\right)}^{0}&& \text{Simplify}. \\ & = 1&& \text{The zero exponent rule} \end{align}[/latex]
- [latex]\begin{align} \frac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}& = \frac{{\left(3\right)}^{5}\cdot {\left({w}^{2}\right)}^{5}}{{\left(6\right)}^{2}\cdot {\left({w}^{-2}\right)}^{2}}&& \text{The power of a product rule} \\ & = \frac{{3}^{5}{w}^{2\cdot 5}}{{6}^{2}{w}^{-2\cdot 2}}&& \text{The power rule} \\ & = \frac{243{w}^{10}}{36{w}^{-4}} && \text{Simplify}. \\ & = \frac{27{w}^{10-\left(-4\right)}}{4}&& \text{The quotient rule and reduce fraction} \\ & = \frac{27{w}^{14}}{4}&& \text{Simplify}. \end{align}[/latex]
Try It
Simplify each expression and write the answer with positive exponents only.- [latex]{\left(2u{v}^{-2}\right)}^{-3}[/latex]
- [latex]{x}^{8}\cdot {x}^{-12}\cdot x[/latex]
- [latex]{\left(\dfrac{{e}^{2}{f}^{-3}}{{f}^{-1}}\right)}^{2}[/latex]
- [latex]\left(9{r}^{-5}{s}^{3}\right)\left(3{r}^{6}{s}^{-4}\right)[/latex]
- [latex]{\left(\frac{4}{9}t{w}^{-2}\right)}^{-3}{\left(\frac{4}{9}t{w}^{-2}\right)}^{3}[/latex]
- [latex]\dfrac{{\left(2{h}^{2}k\right)}^{4}}{{\left(7{h}^{-1}{k}^{2}\right)}^{2}}[/latex]
Answer:
- [latex]\dfrac{{v}^{6}}{8{u}^{3}}[/latex]
- [latex]\dfrac{1}{{x}^{3}}[/latex]
- [latex]\dfrac{{e}^{4}}{{f}^{4}}[/latex]
- [latex]\dfrac{27r}{s}[/latex]
- [latex]1[/latex]
- [latex]\dfrac{16{h}^{10}}{49}[/latex]
Scientific Notation
Recall at the beginning of the section that we found the number [latex]1.3\times {10}^{13}[/latex] when describing bits of information in digital images. Other extreme numbers include the width of a human hair, which is about 0.00005 m, and the radius of an electron, which is about 0.00000000000047 m. How can we effectively work read, compare, and calculate with numbers such as these? A shorthand method of writing very small and very large numbers is called scientific notation, in which we express numbers in terms of exponents of 10. To write a number in scientific notation, move the decimal point to the right of the first digit in the number. Write the digits as a decimal number between 1 and 10. Count the number of places n that you moved the decimal point. Multiply the decimal number by 10 raised to a power of n. If you moved the decimal left as in a very large number, [latex]n[/latex] is positive. If you moved the decimal right as in a small large number, [latex]n[/latex] is negative. For example, consider the number 2,780,418. Move the decimal left until it is to the right of the first nonzero digit, which is 2. We obtain 2.780418 by moving the decimal point 6 places to the left. Therefore, the exponent of 10 is 6, and it is positive because we moved the decimal point to the left. This is what we should expect for a large number.A General Note: Scientific Notation
A number is written in scientific notation if it is written in the form [latex]a\times {10}^{n}[/latex], where [latex]1\le |a|<10[/latex] and [latex]n[/latex] is an integer.Example: Converting Standard Notation to Scientific Notation
Write each number in scientific notation.- Distance to Andromeda Galaxy from Earth: 24,000,000,000,000,000,000,000 m
- Diameter of Andromeda Galaxy: 1,300,000,000,000,000,000,000 m
- Number of stars in Andromeda Galaxy: 1,000,000,000,000
- Diameter of electron: 0.00000000000094 m
- Probability of being struck by lightning in any single year: 0.00000143
Answer:
1. [latex]\begin{align}&\underset{\leftarrow 22\text{ places}}{{24,000,000,000,000,000,000,000\text{ m}}} \\ &2.4\times {10}^{22}\text{ m} \\ \text{ } \end{align}[/latex]
2. [latex]\begin{align}&\underset{\leftarrow 21\text{ places}}{{1,300,000,000,000,000,000,000\text{ m}}} \\ &1.3\times {10}^{21}\text{ m} \\ &\text{ } \end{align}[/latex]
3. [latex]\begin{align}&\underset{\leftarrow 12\text{ places}}{{1,000,000,000,000}} \\ &1\times {10}^{12} \\ \text{ }\end{align}[/latex]
4. [latex]\begin{align}&\underset{\rightarrow 6\text{ places}}{{0.00000000000094\text{ m}}} \\ &9.4\times {10}^{-13}\text{ m} \\ \text{ }\end{align}[/latex]
5. [latex]\begin{align}\underset{\to 6\text{ places}}{{0.00000143}} \\ 1.43\times {10}^{-6} \\ \text{ }\end{align}[/latex]
Analysis of the Solution
Observe that, if the given number is greater than 1, as in examples a–c, the exponent of 10 is positive; and if the number is less than 1, as in examples d–e, the exponent is negative.Try It
Write each number in scientific notation.- U.S. national debt per taxpayer (April 2014): $152,000
- World population (April 2014): 7,158,000,000
- World gross national income (April 2014): $85,500,000,000,000
- Time for light to travel 1 m: 0.00000000334 s
- Probability of winning lottery (match 6 of 49 possible numbers): 0.0000000715
Answer:
- [latex]$1.52\times {10}^{5}[/latex]
- [latex]7.158\times {10}^{9}[/latex]
- [latex]$8.55\times {10}^{13}[/latex]
- [latex]3.34\times {10}^{-9}[/latex]
- [latex]7.15\times {10}^{-8}[/latex]
Converting from Scientific to Standard Notation
To convert a number in scientific notation to standard notation, simply reverse the process. Move the decimal [latex]n[/latex] places to the right if [latex]n[/latex] is positive or [latex]n[/latex] places to the left if [latex]n[/latex] is negative and add zeros as needed. Remember, if [latex]n[/latex] is positive, the value of the number is greater than 1, and if [latex]n[/latex] is negative, the value of the number is less than one.Example: Converting Scientific Notation to Standard Notation
Convert each number in scientific notation to standard notation.- [latex]3.547\times {10}^{14}[/latex]
- [latex]-2\times {10}^{6}[/latex]
- [latex]7.91\times {10}^{-7}[/latex]
- [latex]-8.05\times {10}^{-12}[/latex]
Answer: 1. [latex-display]\begin{align}&3.547\times {10}^{14} \\ &\underset{\to 14\text{ places}}{{3.54700000000000}} \\ &354,700,000,000,000 \\ \text{ }\end{align}[/latex-display] 2. [latex-display]\begin{align}&-2\times {10}^{6} \\ &\underset{\to 6\text{ places}}{{-2.000000}} \\ &-2,000,000 \\ \text{ }\end{align}[/latex-display] 3. [latex-display]\begin{align}&7.91\times {10}^{-7} \\ &\underset{\to 7\text{ places}}{{0000007.91}} \\ &0.000000791 \\ \text{ }\end{align}[/latex-display] 4. [latex-display]\begin{align}&-8.05\times {10}^{-12} \\ &\underset{\to 12\text{ places}}{{-000000000008.05}} \\ &-0.00000000000805 \\ \text{ }\end{align}[/latex-display]
Try It
Convert each number in scientific notation to standard notation.- [latex]7.03\times {10}^{5}[/latex]
- [latex]-8.16\times {10}^{11}[/latex]
- [latex]-3.9\times {10}^{-13}[/latex]
- [latex]8\times {10}^{-6}[/latex]
Answer:
- [latex]703,000[/latex]
- [latex]-816,000,000,000[/latex]
- [latex]-0.00000000000039[/latex]
- [latex]0.000008[/latex]
Using Scientific Notation in Applications
Scientific notation, used with the rules of exponents, makes calculating with large or small numbers much easier than doing so using standard notation. For example, suppose we are asked to calculate the number of atoms in 1 L of water. Each water molecule contains 3 atoms (2 hydrogen and 1 oxygen). The average drop of water contains around [latex]1.32\times {10}^{21}[/latex] molecules of water and 1 L of water holds about [latex]1.22\times {10}^{4}[/latex] average drops. Therefore, there are approximately [latex]3\cdot \left(1.32\times {10}^{21}\right)\cdot \left(1.22\times {10}^{4}\right)\approx 4.83\times {10}^{25}[/latex] atoms in 1 L of water. We simply multiply the decimal terms and add the exponents. Imagine having to perform the calculation without using scientific notation! When performing calculations with scientific notation, be sure to write the answer in proper scientific notation. For example, consider the product [latex]\left(7\times {10}^{4}\right)\cdot \left(5\times {10}^{6}\right)=35\times {10}^{10}[/latex]. The answer is not in proper scientific notation because 35 is greater than 10. Consider 35 as [latex]3.5\times 10[/latex]. That adds a ten to the exponent of the answer.Example: Using Scientific Notation
Perform the operations and write the answer in scientific notation.- [latex]\left(8.14\times {10}^{-7}\right)\left(6.5\times {10}^{10}\right)[/latex]
- [latex]\left(4\times {10}^{5}\right)\div \left(-1.52\times {10}^{9}\right)[/latex]
- [latex]\left(2.7\times {10}^{5}\right)\left(6.04\times {10}^{13}\right)[/latex]
- [latex]\left(1.2\times {10}^{8}\right)\div \left(9.6\times {10}^{5}\right)[/latex]
- [latex]\left(3.33\times {10}^{4}\right)\left(-1.05\times {10}^{7}\right)\left(5.62\times {10}^{5}\right)[/latex]
Answer: 1. [latex-display]\begin{align}\left(8.14 \times 10^{-7}\right)\left(6.5 \times 10^{10}\right) & =\left(8.14 \times 6.5\right)\left(10^{-7} \times 10^{10}\right) && \text{Commutative and associative properties of multiplication} \\ & =\left(52.91\right)\left(10^{3}\right) && \text{Product rule of exponents} \\ & =5.291 \times 10^{4} && \text{Scientific notation} \\ \text{ } \end{align}[/latex-display] 2. [latex-display]\begin{align} \left(4\times {10}^{5}\right)\div \left(-1.52\times {10}^{9}\right)& = \left(\frac{4}{-1.52}\right)\left(\frac{{10}^{5}}{{10}^{9}}\right)&& \text{Commutative and associative properties of multiplication} \\ & = \left(-2.63\right)\left({10}^{-4}\right)&& \text{Quotient rule of exponents} \\ & = -2.63\times {10}^{-4}&& \text{Scientific notation} \\ \text{ } \end{align}[/latex-display] 3. [latex-display]\begin{align} \left(2.7\times {10}^{5}\right)\left(6.04\times {10}^{13}\right)& = \left(2.7\times 6.04\right)\left({10}^{5}\times {10}^{13}\right)&& \text{Commutative and associative properties of multiplication} \\ & = \left(16.308\right)\left({10}^{18}\right)&& \text{Product rule of exponents} \\ & = 1.6308\times {10}^{19}&& \text{Scientific notation} \\ \text{ } \end{align}[/latex-display] 4. [latex-display]\begin{align} \left(1.2\times {10}^{8}\right)\div \left(9.6\times {10}^{5}\right)& = \left(\frac{1.2}{9.6}\right)\left(\frac{{10}^{8}}{{10}^{5}}\right)&& \text{Commutative and associative properties of multiplication} \\ & = \left(0.125\right)\left({10}^{3}\right)&& \text{Quotient rule of exponents} \\ & = 1.25\times {10}^{2}&& \text{Scientific notation} \\ \text{ } \end{align}[/latex-display] 5. [latex-display]\begin{align} \left(3.33\times {10}^{4}\right)\left(-1.05\times {10}^{7}\right)\left(5.62\times {10}^{5}\right)& = \left[3.33\times \left(-1.05\right)\times 5.62\right]\left({10}^{4}\times {10}^{7}\times {10}^{5}\right) \\ & \approx \left(-19.65\right)\left({10}^{16}\right) \\ & = -1.965\times {10}^{17} \end{align}[/latex-display]
Try It
Perform the operations and write the answer in scientific notation.- [latex]\left(-7.5\times {10}^{8}\right)\left(1.13\times {10}^{-2}\right)[/latex]
- [latex]\left(1.24\times {10}^{11}\right)\div \left(1.55\times {10}^{18}\right)[/latex]
- [latex]\left(3.72\times {10}^{9}\right)\left(8\times {10}^{3}\right)[/latex]
- [latex]\left(9.933\times {10}^{23}\right)\div \left(-2.31\times {10}^{17}\right)[/latex]
- [latex]\left(-6.04\times {10}^{9}\right)\left(7.3\times {10}^{2}\right)\left(-2.81\times {10}^{2}\right)[/latex]
Answer:
- [latex]-8.475\times {10}^{6}[/latex]
- [latex]8\times {10}^{-8}[/latex]
- [latex]2.976\times {10}^{13}[/latex]
- [latex]-4.3\times {10}^{6}[/latex]
- [latex]\approx 1.24\times {10}^{15}[/latex]
Example: Applying Scientific Notation to Solve Problems
In April 2014, the population of the United States was about 308,000,000 people. The national debt was about $17,547,000,000,000. Write each number in scientific notation, rounding figures to two decimal places, and find the amount of the debt per U.S. citizen. Write the answer in both scientific and standard notations.Answer: The population was [latex]308,000,000=3.08\times {10}^{8}[/latex]. The national debt was [latex]\$ 17,547,000,000,000 \approx \$1.75 \times 10^{13}[/latex]. To find the amount of debt per citizen, divide the national debt by the number of citizens.
Try It
An average human body contains around 30,000,000,000,000 red blood cells. Each cell measures approximately 0.000008 m long. Write each number in scientific notation and find the total length if the cells were laid end-to-end. Write the answer in both scientific and standard notations.Answer: Number of cells: [latex]3\times {10}^{13}[/latex]; length of a cell: [latex]8\times {10}^{-6}[/latex] m; total length: [latex]2.4\times {10}^{8}[/latex] m or [latex]240,000,000[/latex] m.
Key Equations
Rules of Exponents For nonzero real numbers [latex]a[/latex] and [latex]b[/latex] and integers [latex]m[/latex] and [latex]n[/latex] | |
Product rule | [latex]{a}^{m}\cdot {a}^{n}={a}^{m+n}[/latex] |
Quotient rule | [latex]\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}[/latex] |
Power rule | [latex]{\left({a}^{m}\right)}^{n}={a}^{m\cdot n}[/latex] |
Zero exponent rule | [latex]{a}^{0}=1[/latex] |
Negative rule | [latex]{a}^{-n}=\dfrac{1}{{a}^{n}}[/latex] |
Power of a product rule | [latex]{\left(a\cdot b\right)}^{n}={a}^{n}\cdot {b}^{n}[/latex] |
Power of a quotient rule | [latex]{\left(\dfrac{a}{b}\right)}^{n}=\dfrac{{a}^{n}}{{b}^{n}}[/latex] |
Key Concepts
- Products of exponential expressions with the same base can be simplified by adding exponents.
- Quotients of exponential expressions with the same base can be simplified by subtracting exponents.
- Powers of exponential expressions with the same base can be simplified by multiplying exponents.
- An expression with exponent zero is defined as 1.
- An expression with a negative exponent is defined as a reciprocal.
- The power of a product of factors is the same as the product of the powers of the same factors.
- The power of a quotient of factors is the same as the quotient of the powers of the same factors.
- The rules for exponential expressions can be combined to simplify more complicated expressions.
- Scientific notation uses powers of 10 to simplify very large or very small numbers.
- Scientific notation may be used to simplify calculations with very large or very small numbers.
Glossary
scientific notation a shorthand notation for writing very large or very small numbers in the form [latex]a\times {10}^{n}[/latex] where [latex]1\le |a|<10[/latex] and [latex]n[/latex] is an integerLicenses & Attributions
CC licensed content, Original
- Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
CC licensed content, Shared previously
- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
- Product Rule for Exponents. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Quotient Rule for Exponents. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Using the Power Rule to Simplify Expressions With Exponents. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Question ID 93370, 93399, 93402, 93393. Authored by: Michael Jenck. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- Question ID 1961, 2874. Authored by: David Lippman. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- Simplify Expressions With Zero Exponents. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Simplify Expressions With Negative Exponents. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Power of a Product. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Power of a Quotient. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Question ID 44120, 43231. Authored by: Brenda Gardner. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- Question ID 7833, 14060. Authored by: Tyler Wallace. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- Question ID 109762, 109765. Authored by: Lumen Learning. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- Question ID 51959, 101856, 102452. Authored by: Roy Shahbazian. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- Question ID 14047, 14058, 14059, 14046, 14051, 14056, 14057.. Authored by: James Sousa. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- Question ID 43896. Authored by: Carla Kulinsky. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- Question ID 2466. Authored by: Bryan Jones. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- Question ID 3295. Authored by: Norm Friehauf. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
CC licensed content, Specific attribution
- College Algebra. Provided by: OpenStax Authored by: OpenStax College Algebra. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.